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A Theory of Isolatability Scott Aaronson Andrew Drucker MIT Freeze-Dried Computation Motivating Question: How much useful computational work can one “store” in (say) an n-qubit quantum state, or a coin whose bias is an arbitrary real number? Potentially a huge amount! We give a new tool—called “isolatability”—for ruling out the possibility of such extravagant encodings. Idea: Take some advice resource (such as a coin or a quantum state), and simulate it using a short classical string, together with a polynomial number of untrusted advice resources. In other words, all the relevant information in the advice resource gets packed into an ordinary string (which we say “isolates” the resource), and we're left with just a computational search problem—of finding coins, quantum states, etc. that are consistent with the string. Part I: The Majority-Certificates Lemma Our basic tool Part II: Application to Quantum Advice BQP/qpoly QMA/poly Part III: Application to Advice Coins PSPACE/coin PSPACE/poly Part I: The Majority- Certificates Lemma Intuition: We’re given a black box x f f(x) that computes some Boolean function f:{0,1}n{0,1} belonging to a “small” set S (meaning, of size 2poly(n)). Someone wants to prove to us that f equals (say) the all-0 function, by having us check a polynomial number of outputs f(x1),…,f(xm). This is trivially impossible! f0 f1 f2 f3 f4 f5 But what if we get 3 black x1 0 1 0 0 0 0 boxes, and are allowed to x2 0 0 1 0 0 0 simulate f=f0 by taking the x3 0 0 0 1 0 0 point-wise MAJORITY of x4 0 0 0 0 1 0 their outputs? x5 0 0 0 0 0 1 Majority-Certificates Lemma Definitions: A certificate is a partial Boolean function C:{0,1}n{0,1,*}. A Boolean function f:{0,1}n{0,1} is consistent with C, if f(x)=C(x) whenever C(x){0,1}. The size of C is the number of inputs x such that C(x){0,1}. Lemma: Let S be a set of Boolean functions f:{0,1}n{0,1}, and let f*S. Then there exist m=O(n) certificates C1,…,Cm, each of size O(log|S|), such that (i) Some fiS is consistent with each Ci, and (ii) If f1S is consistent with C1, f2S is consistent with C2, and so on, then MAJ(f1,…,fm)=f*, where MAJ denotes pointwise majority. Proof Idea By symmetry, we can assume f* is the all-0 function. Consider a two-player, zero-sum matrix game: Bob picks an input x{0,1}n The lemma follows from this claim! Just choose certificates C1,…,Cm independently from Alice’s winning Alice picks a certificate by a Chernoff bound, almost certainly distribution. Then MAJ(f1 consistent C of size k(x),…,fm(x))=0 for all f1,…,fm consistent with C1,…,Cm with some fS respectively and all inputs x{0,1}n. So clearly there exist C1,…,Cm with this property. Alice wins this game if f(x)=0 for all fS consistent with C. Crucial Claim: Alice has a mixed strategy that lets her win >90% of the time. Proof of Claim Use the Minimax Theorem! Given a distribution D over x, it’s enough to create a fixed certificate C such that Pr f consistent with C s.t. f x 1 . 1 xD 10 Stage I: Choose x1,…,xt independently from D, for some t=O(log|S|). Then with high probability, requiring f(x1)=…=f(xt)=0 kills off every fS such that Pr f x 1 . 1 xD 10 Stage II: Repeatedly add a constraint f(xi)=bi that kills at least half the remaining functions. After ≤ log2|S| iterations, we’ll have winnowed S down to just a single function fS. Part II: Application to Quantum Advice BQP/qpoly is the class of problems solvable in quantum polynomial time, with the help of polynomial-size “quantum advice states” Formally: a language L is in BQP/qpoly if there exists a polynomial time quantum algorithm A, as well as quantum advice states {|n}n on poly(n) qubits, such that for every input x of size n, A(x,|n) decides whether or not xL with error probability at most 1/3 YQP (“Yoda Quantum Polynomial-Time”) is the same, except we also require that for every alleged advice state , A(x,) outputs either the right answer or “FAIL” with probability at least 2/3 BQP YQP QMA BQP/qpoly QUANTUM ADVICE IS POWERFUL Watrous 2000: For any fixed, finite black-box group Gn and subgroup Hn≤Gn, deciding membership in Hn is in BQP/qpoly The quantum advice state is just an equal superposition |Hn over the elements of Hn We don’t know how to solve the same problem in BQP/poly A.-Kuperberg 2007: There exists a “quantum oracle” separating BQP/qpoly from BQP/poly NO IT ISN’T A. 2004: BQP/qpoly PP/poly = PostBQP/poly Quantum advice can be simulated by classical advice, combined with postselection on unlikely measurement outcomes A. 2006: HeurBQP/qpoly = HeurYQP/poly Trusted quantum advice can be simulated on most inputs by trusted classical advice combined with untrusted quantum advice New Result: BQP/qpoly = YQP/poly Trusted quantum advice is equivalent in power to trusted classical advice combined with untrusted quantum advice. (“Quantum states never need to be trusted”) “PHYSICS” IMPLICATION Let ρ be any quantum state on n qubits. Then for all m,ε, there exists a 2-local Hamiltonian H=H1+⋯+HL on poly(n,m,1/ε) qubits, such that any ground state |φ of H can be used to simulate ρ (with error ε) on all quantum circuits of size at most m. In other words, there exists an efficient mapping C→C′ such that for all circuits C of size m, PrC ' accepts PrC accepts . What Does It Mean? Preparing quantum advice states is no harder than preparing ground states of local Hamiltonians This explains a once-mysterious relationship between quantum proofs and quantum advice: efficient preparability of ground states would imply both QMA=QCMA and BQP/qpoly=BQP/poly “Quantum Karp-Lipton Theorem”: NP-complete problems are not efficiently solvable using quantum advice, unless some uniform complexity classes collapse unexpectedly QCMA/qpoly QMA/poly: classical proofs and quantum advice can be simulated with quantum proofs and classical advice PSPACE/poly A.’06 QMA/qpoly PP/poly QMA/poly PP This work QCMA/qpoly BQP/qpoly QCMA/poly QMA =YQP/poly BQP/poly YQP QCMA BQP Holevo’s Theorem Minimax Circuit Learning Covering Lemma Theorem (Bshouty et al.) (Alon et al.) Random Access Code Lower Bound Learning of p- (Ambainis et al.) Majority- Safe Concept Classes Certificates Winnowing (Bartlett & Long) Fat-Shattering Bound Lemma Lemma (A.’06) QMA=QMA+ Real Majority- (Aharonov & Regev) Certificates Lemma Cook-Levin Theorem HeurBQP/qpoly=HeurYQP/poly (A.’06) LOCAL HAMILTONIANS is BQP/qpoly=YQP/poly QMA-complete (Kitaev) Used as lemma Quantum advice no harder than ground state preparation Generalizes “Lifting” the Majority-Certificates Lemma Boolean Majority-Certificates BQP/qpoly=YQP/poly Proof Set S of Boolean functions Set S of p(n)-qubit mixed states “True” function f*S “True” advice state |n Other functions f1,…,fm Other states 1,…,m Certificate Ci to isolate fi Measurement Ei to isolate I New Difficulty Solution The class of p(n)-qubit quantum states is Result of A.’06 on learnability of quantum infinitely large! And even if we discretize it, it’s states (building on Ambainis et al. 1999) still doubly-exponentially large Instead of Boolean functions f:{0,1}n{0,1}, Learning theory has tools to deal with now we have real functions f:{0,1}n[0,1] this: fat-shattering dimension, -covers… representing the expectation values (Alon et al. 1997) How do we verify a quantum witness without QMA=QMA+ (Aharonov & Regev 2003) destroying it? What if a certificate asks us to verify Tr(E)≤a, “Safe Winnowing Lemma” but Tr(E) is “right at the knife-edge”? Majority-Certificates Lemma, Real Case Lemma: Let S be a set of functions f:{0,1}ⁿ→[0,1], let f∗∈S, and let ε>0. Then we can find m=O(n/ε²) functions f1,…,fm∈S, sets X1,…,Xm⊆{0,1}ⁿ each of size n and 2 k O 3 fat / 48 S , n fat S / 48 for which the following holds. All functions g1,…,gm∈S that satisfy max g i x f i x for all i[m] also satisfy xX i maxn g x f * x , x0 ,1 g x : g1 x g m x . where 1 m Theorem: BQP/qpoly = YQP/poly. Proof Sketch: YQP/poly BQP/qpoly is immediate. For the other direction, let LBQP/qpoly. Let M be a quantum algorithm that decides L using advice state |n. Define f x : Pr M x, accepts Let S = {f : }. Then S has “fat-shattering dimension” at most poly(n), by A.’06. So we can apply the real version of the Majority-Certificates Lemma to S. This yields certificates C1,…,Cm (for some m=poly(n)), such that any states 1,…,m consistent with C1,…,Cm respectively satisfy 1 m f 1 x f m x f n n x for all x{0,1}n (regardless of entanglement). To check the Ci’s, we use the “QMA+ super-verifier” of Aharonov & Regev. Quantum Karp-Lipton Theorem Karp-Lipton 1982: If NP P/poly, then coNPNP = NPNP. Our quantum analogue: If NP BQP/qpoly, then coNPNP QMAPromiseQMA. Proof Idea: A coNPNP statement has the form x y R(x,y). By the hypothesis and BQP/qpoly = YQP/poly, there exists an advice string s, such that any quantum state consistent with s lets us solve NP problems (and some such is consistent). In QMAPromiseQMA, first guess an s that’s consistent with some state . Then use the oracle to search for an x and such that, if is consistent with s, then R(x,Q(x,)) holds, where Q is a quantum algorithm that searches for a y such that R(x,y). Part III: Application to Advice Coins Erik Demaine (motivated by a computational genetics problem): “Suppose a PSPACE machine can flip a coin with Bernoulli probability p an unlimited number of times. Can it extract an exponential amount of information (or even more) about p?” Me: “I’m sure whatever the answer is, it’s obvious...” Didn’t seem too likely there could be superpowerful “Advice Coins” Indeed, Hellman & Cover proved the following in 1970... Suppose a finite automaton M is trying to decide whether a coin has p=½ or p=½+. Then even if it can flip the coin an unlimited number of times, M needs (1/) states to succeed with probability (say) 2/3. This result seems to imply PSPACE/coinPSPACE/poly, since we could take the first poly(n) bits of p as the advice. But it breaks down if p is close to 0 or 1! Furthermore, quantum mechanics nullifies the Hellman-Cover Theorem! Theorem: For any >0, it’s possible to distinguish a coin with p=½ from a coin with p=½+ using a single qubit of memory, with error probability independent of . 1 Halt with probability ~2/100 at each time step the coin. Keep flipping 0 1 angle after halting, in coin Expected difference in final 2 Whenever the p=½ vs. p=½+ cases: 1 radian lands heads, rotate Standard deviation in angle: /100 radians 0 counterclockwise. Whenever it lands tails, rotate /100 radians clockwise. Theorem: Despite these obstacles, BQPSPACE/coin = PSPACE/coin = PSPACE/poly. Proof: Suffices to show BQPSPACE/coin PSPACE/poly. Let 0 = quantum operation applied to our memory qubits whenever coin lands heads, 1 = operation applied when it lands tails Then induced operation at each time step: We’re interested in a fixed-point of p: a mixed state p such that Fixed-Points of Superoperators Already studied by [A.-Watrous 2008], in the context of quantum computing with closed timelike curves Our result there: BQPCTC = PCTC = PSPACE Quantum computers with CTCs have exactly the same power as classical computers with CTCs, namely PSPACE (or: “CTCs make time and space equivalent as computational resources”) Key Lemma: Let a(p) be the probability that an S-state quantum finite automaton accepts, if each input bit is 1 with independent probability p. Then a(p) is a degree-S2 rational function of p. Proof Idea: We can write a(p) as 1 T lim Acc p 0 Acc Acc p Acc . t T T t 1 Furthermore, each entry of p can be written as a degree-S2 rational function of p, by using Cramer’s Rule on S2S2 matrices. Hence a(p) is itself a degree-S2 rational function of p, except possibly at a finite number of singularities: ax(p) p A further continuity argument rules out the singularities, except possibly at p=0 and p=1. Now, by calculus, a degree-S2 rational function can cross the line y=½ at most 2S2 times… Given a BQPSPACE/coin machine M, let ax(p) be its acceptance probability on input x{0,1}n and a coin with Bernoulli probability p. Challenge: How can we describe p well enough to compute ax(p) for every x, using only poly(n) bits? ax(p) p Finishing the Argument (Sketch) We’ve upper-bounded how many distinct Boolean functions f:{0,1}n{0,1} can be expressed, as we vary p from 0 to 1. So by the Majority-Certificates Lemma, we can simulate PSPACE/coin using a PSPACE/poly machine, combined with poly(n) untrusted advice coins. We then get a computational search problem: finding coin biases p1,…,pm that are consistent with the /poly advice string. We can solve that problem in PSPACE, using NC algorithms for root-finding developed by Neff and Pan When Can An O(1)-State Finite Automaton Detect an Change to the Bias of a Coin? Open Problems Find other applications of isolatability Circuit complexity? Communication complexity? Learning theory? Quantum information? Optimality of the Majority-Certificates Lemma? Quantum finite automata: are their limiting acceptance probabilities continuous functions of p, for p(0,1)? Prove a classical oracle separation between BQP/poly and BQP/qpoly=YQP/poly Promised Application to “Physics” By Kitaev et al., we know LOCAL HAMILTONIANS is QMA-complete. Furthermore, in their reduction, the witness is a “history state” T 1 : T t t 1 t Measuring this state yields the original QMA witness |1 with (1/poly(n)) probability. Hence |1 can be recovered from poly n ' : So given any language LBQP/qpoly=YQP/poly, we can use the Kitaev et al. reduction to get a local Hamiltonian H whose unique ground state is |’. We can then use |’ to recover the YQP witness |, and thereby decide L

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